Find the General Anti-derivative (Calculus I)

[treehau5]

Homework Statement

given f(x) = [x^3+5sqrt(x)]/x^2, find the anti-derivative

Homework Equations

The Attempt at a Solution

Hi I have attempted to solve this by re-writing the equation as a sum of two fractions:

x^3/x^2 + 5sqrt(x)/x^2, simplying gives = x + 5/x^3/2

I then apply the principle anti-differentiation rules, and I come out with:

x^2/2 + 5x^-1/2 = x^2/2 + 10/sqrt(x) + C

This answer is coming back in WebWork as wrong. I also checked my answer on wolframalpha, and got the same result.

What am I doing wrong? (If anything)

 

[DryRun]

x^3/x^2 + 5sqrt(x)/x^2, simplying gives = x + 5/x^3/2

Try re-writing it: $$x + \frac{5}{x^{3/2}}=x+5x^{-3/2}$$Then integrate.

 

[treehau5]

I am not quite at intergration yet, this is the last section of my Calculus I course, Calculus II is intergration. So for now our anti-d's are pretty simple and just follow some basic rules.

 

[treehau5]

Ok so
x + 5x^-3/2 following the rule: x(n+1)/(n+1) gives

x¹⁺¹ / (1+1) + 5x^{-3/2 + 2/2} / -(1/2) =

x²/2 + 5x^-1/2/ -(1/2 ) = -10x^-1/2 or finally

x² / 2 + 10 / sqrt(x)

I am still doing it wrong?

 

[treehau5]

ok yes I am doing it wrong, its - 10 / sqrt(x).

Thank you very much.

Story of my life, I always miss a sign.

 

[Mark44]

I am not quite at intergration yet, this is the last section of my Calculus I course, Calculus II is intergration. So for now our anti-d's are pretty simple and just follow some basic rules.

Integration is finding the antiderivative.

Note that there is no such word in English as intergration.

 

[treehau5]

Integration is finding the antiderivative.

Note that there is no such word in English as intergration.

Thank you Mark for the insight. And you are right, there is no such word as intergration.